Compact null sections imply a compact set is null

If $a=b$, compactness bounds $K$ inside $\{a\}\times [-R,R{]}^n$ for some $R>0$. Divide the $n$-dimensional base into a finite grid of cubes of side at most $\delta$ and cover each corresponding point of $K$ by an $(n+1)$-cube of side $\delta$ centred in the first coordinate at $a$. The number of grid cubes is at most $(2R\mathrm{/}\delta +2{)}^n$, so their total $(n+1)$-volume is at most $(2R+2\delta {)}^n\delta \to 0$. By [F1], $K$ is null. Assume henceforth that $a<b$.

Fix $\epsilon >0$ and put $L:=b-a+1$. For each $t\in [a,b]$, [F1] supplies the following covers. [F1, given, choose] There are finitely many closed $n$-cubes $Q_{t,1},\dots ,Q_{t,m_t}$ covering the compact section $K_t$ with total $n$-volume below $\epsilon \mathrm{/}(2^{n+3}L)$. Enlarge them slightly to open $n$-cubes ${\tilde{Q}}_{t,\mathrm{\ell }}\supseteq Q_{t,\mathrm{\ell }}$ so that, with $O_t:={\bigcup }_{\mathrm{\ell }=1}^{m_t}{\tilde{Q}}_{t,\mathrm{\ell }},$ one still has ${\sum }_{\mathrm{\ell }=1}^{m_t}{\mathrm{vol}}_n({\tilde{Q}}_{t,\mathrm{\ell }})<\frac{\epsilon }{2^{n+2}L}\mathrm{.}$ Thus each section has an open finite cube cover with the stated uniform volume budget.

Since $K_t\subseteq O_t$ and $O_t$ is open, compactness of $K$ gives the following interval. [step 1.2, given, contradiction] There is an open interval $I_t$ about $t$ such that $K\cap (I_t\times {\mathbb{R}}^n)\subseteq I_t\times O_t\mathrm{.}$ Otherwise one could find $(t_j,y_j)\in K$ with $t_j\to t$ and $y_j\mathrm{\notin }{O}_t$; passing to a convergent subsequence inside the compact set $K$ yields a limit point $(t,y)\in K$ with $y\in K_t\setminus O_t$, contradiction.

By [L1], the cover $\{I_t:t\in [a,b]\}$ has a Lebesgue number. Choose a finite partition of $[a,b]$ into closed intervals $J_1,\dots ,J_r$ of positive length smaller than that number, and for each $j$ choose $t_j$ with $J_j\subseteq I_{t_j}$. Then ${\sum }_j\mathrm{\mid }{J}_j\mathrm{\mid }=b-a<L$. For each prism $J_j\times {\tilde{Q}}_{t_j,\mathrm{\ell }}$, use the following subdivision. [L1, step 2.1, choose] Let ${\lambda }_j:=\mathrm{\mid }{J}_j\mathrm{\mid }$ and let $s_{j,\mathrm{\ell }}$ be the side length of ${\tilde{Q}}_{t_j,\mathrm{\ell }}$. If ${\lambda }_j\ge s_{j,\mathrm{\ell }}$, partition the interval direction into at most $\lceil {\lambda }_j\mathrm{/}{s}_{j,\mathrm{\ell }}\rceil$ pieces of length at most $s_{j,\mathrm{\ell }}$; if ${\lambda }_j\le s_{j,\mathrm{\ell }}$, partition each of the $n$ base directions into at most $\lceil s_{j,\mathrm{\ell }}\mathrm{/}{\lambda }_j\rceil$ pieces of length at most ${\lambda }_j$. In either case the prism is covered by finitely many closed $(n+1)$-cubes of total volume at most $2^n{\lambda }_j{s}_{j,\mathrm{\ell }}^n=2^n\mathrm{\mid }{J}_j\mathrm{\mid }{\mathrm{vol}}_n({\tilde{Q}}_{t_j,\mathrm{\ell }})\mathrm{.}$ These cubes cover $K$, and their total volume is at most $2^n{\sum }_{j=1}^r\mathrm{\mid }{J}_j\mathrm{\mid }{\sum }_{\mathrm{\ell }=1}^{m_{t_j}}{\mathrm{vol}}_n({\tilde{Q}}_{t_j,\mathrm{\ell }})<2^{n}L\cdot \frac{\epsilon }{2^{n+2}L}<\epsilon \mathrm{.}$ By [F1], $K$ is null.

Therefore compact null sections imply the whole compact set is null.